On Mar 9, 2:03pm, S. Cowles wrote:
} Subject: parameter expansion, substitution
}
} i am not getting expected results in the following code snippet:
} b="abc" ; b=${b/#abc%/d} ; echo ${b}
} i expect the value of b to be changed to "d"; however, it remains "abc".
You've misunderstood. You want:
b="abc" ; b=${b/#%abc/d} ; echo ${b}
The # or % or #% must appear at the beginning of the pattern, they are
not positional like ^ and $ in a regular expression.